Question

Let $z=x+iy$, then find the locus of $z$ for $|z+1{|}^2-|z-1{|}^2=4$ is

• A. $x^2+y^2=1$
• B. $x=1$
• C. $y=0$
• D. none of these

Answer 1

The correct option is B $x=1$

We have,

$z=x+iy$ ……. (1)

Since,

${|z+1|}^2-{|z-1|}^2=4$

On putting the value of the $z$, we get

${|x+iy+1|}^2-{|x+iy-1|}^2=4$

${|x+1+iy|}^2-{|x-1+iy|}^2=4$

${(x+1)}^2+y^2-[{(x-1)}^2+y^2]=4$

${(x+1)}^2+y^2-{(x-1)}^2-y^2=4$

$x^2+1+2x-x^2-1+2x=4$

$4x=4$

$x=1$

Hence, this is the answer.